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\(S=-\frac{1}{2}\left(3x+3-2\sqrt{2x^2+5x+2}+x+7-4\sqrt{x+3}\right)+5\)
\(=-\frac{1}{2}\left[\frac{\left(x-1\right)^2}{3x+3+2\sqrt{2x^2+5x+2}}+\frac{\left(x-1\right)^2}{x+7+4\sqrt{x+3}}\right]+5\le5\)
\(S_{max}=5\) khi \(x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT cosi:
\(A=\sqrt{\left(2x+1\right)\left(x+2\right)}+2\sqrt{x+3}-2x\\ A\le\dfrac{2x+1+x+2}{2}+\dfrac{4+x+3}{2}-2x\\ A\le\dfrac{3x+3}{2}+\dfrac{x+7}{2}-2x=\dfrac{3x+3+x+7-4x}{2}=5\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}2x+1=x+2\\4=x+3\end{matrix}\right.\Leftrightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\sqrt{2x^2+5x+2}+2\sqrt{x+3}-2x\)
\(2A=2\sqrt{2x^2+5x+2}+4\sqrt{x+3}-4x\)
\(2A=2\sqrt{\left(2x+1\right)\left(x+2\right)}+4\sqrt{x+3}-4x\)
\(\le2x+1+x+2+4+x+3-4x=10\)
=>2A\(\le10\Rightarrow A\le5\)
dấu bằng xảy ra \(\Leftrightarrow2x+1=x+2\)
và x+3=4
=>x=1
maxA=5 khi x=1
![](https://rs.olm.vn/images/avt/0.png?1311)